## Thomas' Calculus 13th Edition

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We know that $div F=\dfrac{\partial P}{\partial x}i+\dfrac{\partial Q}{\partial y}j$ So, $div \space F=\dfrac{\partial}{\partial x}[\dfrac{-y}{(x^2+y^2)^{1/2}}]+\dfrac{\partial}{\partial y} [\dfrac{x}{(x^2+y^2)^{1/2}}]$ and $div \space F=\dfrac{2xy}{2}(x^2+y^2)^{-3/2}-\dfrac{2yx}{2}(x^2+y^2)^{-3/2}=0$